SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

22 CHAPTER 3. BASICS
34. (a)
Therefore, (1) = F (2) Y (a).
(b) Let T ≡ number of trials until acceptance. On each trial
{
Pr{accept} =Pr V ≤ f }
Y (Z)
= 1 cf Z (Z) c
Since each trial is independent, T has a geometric distribution with γ =1/c.
Therefore, E[T ]= 1 = c. 1/c
(c) Note that f Y (a) is maximized at f Y (1/2) = 1 1.
2
Let f Z (a) =1, 0 ≤ a ≤ 1
c =1 1 2
Therefore, cf Z (a) =1 1 ≥ f 2 Y (a)
1. U ← random()
2. Z ← U
3. V ← random()
4. if V ≤ 6Z(1 − Z)/(1 1 2 )then
return Y ← Z
else
goto step 1
endif
E[T ]=1 1 2
F (a) =
∫ a
α
2(b − α)
(β − α) 2 db
=
U =
(a − α)2
(β − α) 2 , α ≤ a ≤ β
(X − α)2
(β − α) 2
algorithm
1. U ← random()
2. X ← α +(β − α) √ U
3. return X
(X − α) 2 = U(β − α) 2
X = α +
√
U(β − α) 2 = α +(β − α) √ U